Topic: Motion in a vertical circle· Work, Energy and Power

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v_0 as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at po…

A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:

circle with bob at bottom given v0, string slack at P at angle theta above horizontal through O

1. $(\sin\theta)^{1/2}$

2. $\left(\frac{1}{2+3\sin\theta}\right)^{1/2}$

3. $\left(\frac{\cos\theta}{2+3\sin\theta}\right)^{1/2}$

4. $\left(\frac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Answer: 4. $\left(\frac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$

Key idea: the string goes slack when the tension becomes zero, and energy is conserved along the way.

At P (angle $\theta$ above the horizontal through the centre), only the radial component of gravity supplies the centripetal force when $T=0$:

$$mg\sin\theta=\frac{mv^2}{l}\Rightarrow v^2=gl\sin\theta$$

Height of P above the lowest point is $l+l\sin\theta$. Energy conservation:

$$\tfrac12 mv_0^2=\tfrac12 mv^2+mgl(1+\sin\theta)\Rightarrow v_0^2=gl\sin\theta+2gl+2gl\sin\theta=gl(2+3\sin\theta)$$$$\frac{v}{v_0}=\left(\frac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v_0 as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at po…

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